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(...) think that way, though I'll note I passed Calculus). I come up with 2/3 if you switch. A 1/3 chance that you have the right choice initially. The other two doors are an aggragate 2/3 chance. Even though one is eliminated, it is a wrong one, so (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) though I admit it has been a long time since I have taken my probablity and statistics class. (...) Gosh, everyone is seizing on this: I didn't say it was and it has nothing to do with my answers!!! :-) (...) Sigh. Yes, I know that. (...) Yes, (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) If I were to think about it like a sequence of events, logically leading up to the end choice, it would go something like this-- The event starts-- Contestant picks a door -- 1, 2, or 3 If this is all there is to the contest, we have 33 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) See this one for a clearer picture: (URL)No, because he won't use the same algorithm since he then becomes (...) That's not the problem presented, though. According to the problem, it's KNOWN that he ALWAYS reveals a zonk prize door that you (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Thank you for repeating the message I already read and applied to the situation. I did illuminate that it is not a valid Let's Make a Deal scenario, but it's still the one I'm talking about. -->Bruce<-- (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Doh. This is just coming back to haunt me. I should never have included the "door revealed" column. It's entirely irrelevant, and is only serving to confuse. A better list: Prize Choice Switch or Stay? 1 1 stay! 1 2 switch! 1 3 switch! 2 1 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) That is not a cheat. -->Bruce<-- (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) <mucho snippage> (...) Bruce - I see the confusion. You aren't solving the problem as presented: "Now, here's a trickier one: Ever watched the old game show "Let's Make a Deal"? Sometimes, they'd offer people things similar to this scenario. (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Whew. You are way off in your probablity and statistics analysis: if you want to double everything, then you would have to list 112 and 113 twice also. They are *not* the same selection. (...) and 1 1 2 (...) and 1 1 3 (...) (etc. snip) (...) (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) That doesn't make sense at all-- If I choose door 1 and the prize is behind door 1, the host can choose to show me door 2 *or* door 3 - 2 possibilities If I choose door one and the prize is behind door 2, the host can *only* show me door 3 - (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) This really bugged me about all the links Larry sent. They all claimed that there was some sort of "loophole" if you assume that the host has the option to either: - reveal the door you picked, showing a goat (in which case, yes, it's 50/50) - (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) The host does cheat: he always reveals a losing door. That's part of the basic premise. (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Which is why those are left out: there never enter the scenario. Only those that would actually happen are "invoked". And the simulator supports my conclusions. Only if the host "cheats" does the 2/3 option become statistically correct. :-) (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Actually, I explained (albeit briefly) why this isn't correct. You see, you now have 2 listings for "Choose 1, Prize 1", "Coose 2, Prize 2", and "Choose 3, Prize 3", when each deserves *equal* probability to each other possibility, like (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) Got it... Goes into a fairly comprehensible walk-thru, but the point that makes it obvious (as stated) is: " Imagine that there were a million doors. Also, after you have chosen your door; Monty opens all but one of the remaining doors, (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) You're also being incompletely rigorous. :) If you want to do an "all possible outcomes" probability check, you have to add in all the 'door revealed=prize' options which exist for the sake of the probabilities involved, but are never invoked (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Actually, it supports the 50/50. Host will always offers regardless of whether you chose correctly or not (basic premise - choose the "host does not know"): simulator runs 50/50. If it's a screw you scenario, then yes, you switch, but I (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Liking it or not is irrelevant. The probabilities are what they are. And the answer is still 2/3 if you switch, given the premises. No matter whether you like it or not. Go to the site I gave you and read what is written there. (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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In lugnet.off-topic.geek, Bruce Schlickbernd writes: <snip> (...) So, like others suggest about knowing that one of the marbles is red, the chance that the other one is red... the one that we know becomes basically irrelevant to the chance--there (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) I think your analysis is incorrect, precisely because you did *not* list all outcomes. Choose Prize Door Revealed Possible choices 1 1 2 Stay (win). Switch (lose). 1 1 3 Stay (win). Switch (lose). 1 2 3 Stay (lose). Switch (win). 1 3 2 Stay (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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